Problem: Marta makes $90\%$ of the free throws she attempts. She is going to shoot $3$ free throws. Assume that the results of free throws are independent from each other. Let $X$ represent the number of free throws she makes. Find the probability that Marta makes at least $2$ of the $3$ free throws. You may round your answer to the nearest hundredth. $P(X\geq2)=$
Solution: Strategy (without a fancy calculator) The probability that Marta makes at least $2$ of the $3$ free throws is equivalent to the probability that Marta makes $2$ or $3$ free throws. So we can find those probabilities and add them them together to get our answer: $\begin{aligned} P(X\geq2)&=P(\text{makes }2)+P(\text{makes }3) \\\\ &=P(X=2)+P(X=3) \end{aligned}$ Finding $P(X=3)$ For each free throw, we know $P({\text{make}})={90\%}$. To find the probability that Marta makes all $3$ free throws in a row, we can multiply probabilities since we were told to assume independence: $\begin{aligned} P(X=3)&=({0.90})({0.90})({0.90}) \\\\ &=({0.90})^3 \\\\ &=0.729 \end{aligned}$ We'll come back and use this result later. Next, we need to find $P(X=2)$ (the probability that Marta makes $2$ of the $3$ free throws). Finding $P(X=2)$ Making $2$ free throw in $3$ attempts means Marta needs to make $2$ free throws and miss $1$ free throw. For each free throw, we know $P({\text{make}})={90\%}$ and $P({\text{miss}})={10\%}$. Since we are assuming independence, we can multiply probabilities to find the probability of making $2$ attempts followed by missing $1$ attempt — ${\text{S}}$ is a free throw she makes and ${\text{F}}$ is a free throw she misses. $P({\text{SS}}{\text{F}})=({0.9})^2({0.1})=0.081$ This isn't the entire probability though, because there are other ways to make $2$ shots from $3$ attempts (for example, SFS). How many different ways are there? We can use the combination formula: $\begin{aligned} _n\text{C}_k&=\dfrac{n!}{(n-k)!\cdot k!} \\\\ _3\text{C}_2&=\dfrac{3!}{(3-2)!\cdot2!} \\\\ &=\dfrac{3 \cdot \cancel{2 \cdot 1}}{(1) \cdot \cancel{2 \cdot 1} } \\\\ &=3 \end{aligned}$ There are $3$ ways she can make $2$ shots in $3$ attempts. Do they all have the same probability? Each of the $3$ ways has the same probability that we already found: $\begin{aligned} P({\text{SS}}{\text{F}})&=({0.9})^2({0.1})=0.081 \\\\ P({\text{S}}{\text{F}}{\text{S}})&=({0.9})^2({0.1})=0.081 \\\\ P({\text{F}}{\text{SS}})&=({0.9})^2({0.1})=0.081 \end{aligned}$ So we can multiply this probability by $3$ since that is how many ways there are to make $2$ shots in $3$ attempts. $\begin{aligned} P(X=2)&=3(0.9)^2(0.1) \\\\ &=3(0.081) \\\\ &=0.243 \end{aligned}$ Putting it all together Let's return to our original strategy to answer the question: $\begin{aligned} P(X\geq2)&=P(\text{makes }2)+P(\text{makes }3) \\\\ &=P(X=2)+P(X=3) \\\\ &=3(0.9)^2(0.1)+(0.9)^3 \\\\ &=0.729+0.243 \\\\ &=0.972 \end{aligned}$ The answer $P(X\geq2)=0.972\approx0.97$